\documentclass[11pt]{article} \title{A Tutorial on [Co-]Inductive Types in Coq} \author{Eduardo Gim\'enez\thanks{Eduardo.Gimenez@inria.fr}, Pierre Cast\'eran\thanks{Pierre.Casteran@labri.fr}} \date{May 1998 --- \today} \usepackage{multirow} \usepackage{aeguill} %\externaldocument{RefMan-gal.v} %\externaldocument{RefMan-ext.v} %\externaldocument{RefMan-tac.v} %\externaldocument{RefMan-oth} %\externaldocument{RefMan-tus.v} %\externaldocument{RefMan-syn.v} %\externaldocument{Extraction.v} \input{recmacros} \input{coqartmacros} \newcommand{\refmancite}[1]{{}} %\newcommand{\refmancite}[1]{\cite{coqrefman}} %\newcommand{\refmancite}[1]{\cite[#1] {]{coqrefman}} \usepackage[latin1]{inputenc} \usepackage[T1]{fontenc} \usepackage{makeidx} %\usepackage{multind} \usepackage{alltt} \usepackage{verbatim} \usepackage{amssymb} \usepackage{amsmath} \usepackage{theorem} \usepackage[dvips]{epsfig} \usepackage{epic} \usepackage{eepic} \usepackage{ecltree} \usepackage{moreverb} \usepackage{color} \usepackage{pifont} \usepackage{xr} \usepackage{url} \usepackage{alltt} \renewcommand{\familydefault}{ptm} \renewcommand{\seriesdefault}{m} \renewcommand{\shapedefault}{n} \newtheorem{exercise}{Exercise}[section] \makeindex \begin{document} \maketitle \begin{abstract} This document\footnote{The first versions of this document were entirely written by Eduardo Gimenez. Pierre Cast\'eran wrote the 2004 revision.} is an introduction to the definition and use of inductive and co-inductive types in the {\coq} proof environment. It explains how types like natural numbers and infinite streams are defined in {\coq}, and the kind of proof techniques that can be used to reason about them (case analysis, induction, inversion of predicates, co-induction, etc). Each technique is illustrated through an executable and self-contained {\coq} script. \end{abstract} %\RRkeyword{Proof environments, recursive types.} %\makeRT \addtocontents{toc}{\protect \thispagestyle{empty}} \pagenumbering{arabic} \cleardoublepage \tableofcontents \clearpage \section{About this document} This document is an introduction to the definition and use of inductive and co-inductive types in the {\coq} proof environment. It was born from the notes written for the course about the version V5.10 of {\coq}, given by Eduardo Gimenez at the Ecole Normale Sup\'erieure de Lyon in March 1996. This article is a revised and improved version of these notes for the version V8.0 of the system. We assume that the reader has some familiarity with the proofs-as-programs paradigm of Logic \cite{Coquand:metamathematical} and the generalities of the {\coq} system \cite{coqrefman}. You would take a greater advantage of this document if you first read the general tutorial about {\coq} and {\coq}'s FAQ, both available on \cite{coqsite}. A text book \cite{coqart}, accompanied with a lot of examples and exercises \cite{Booksite}, presents a detailed description of the {\coq} system and its underlying formalism: the Calculus of Inductive Construction. Finally, the complete description of {\coq} is given in the reference manual \cite{coqrefman}. Most of the tactics and commands we describe have several options, which we do not present exhaustively. If some script herein uses a non described feature, please refer to the Reference Manual. If you are familiar with other proof environments based on type theory and the LCF style ---like PVS, LEGO, Isabelle, etc--- then you will find not difficulty to guess the unexplained details. The better way to read this document is to start up the {\coq} system, type by yourself the examples and exercises, and observe the behavior of the system. All the examples proposed in this tutorial can be downloaded from the same site as the present document. The tutorial is organised as follows. The next section describes how inductive types are defined in {\coq}, and introduces some useful ones, like natural numbers, the empty type, the propositional equality type, and the logical connectives. Section \ref{CaseAnalysis} explains definitions by pattern-matching and their connection with the principle of case analysis. This principle is the most basic elimination rule associated with inductive or co-inductive types and follows a general scheme that we illustrate for some of the types introduced in Section \ref{Introduction}. Section \ref{CaseTechniques} illustrates the pragmatics of this principle, showing different proof techniques based on it. Section \ref{StructuralInduction} introduces definitions by structural recursion and proofs by induction. Section~\ref{CaseStudy} presents some elaborate techniques about dependent case analysis. Finally, Section \ref{CoInduction} is a brief introduction to co-inductive types --i.e., types containing infinite objects-- and the principle of co-induction. Thanks to Bruno Barras, Yves Bertot, Hugo Herbelin, Jean-Fran\c{c}ois Monin and Michel L\'evy for their help. \subsection*{Lexical conventions} The \texttt{typewriter} font is used to represent text input by the user, while the \textit{italic} font is used to represent the text output by the system as answers. Moreover, the mathematical symbols \coqle{}, \coqdiff, \(\exists\), \(\forall\), \arrow{}, $\rightarrow{}$ \coqor{}, \coqand{}, and \funarrow{} stand for the character strings \citecoq{<=}, \citecoq{<>}, \citecoq{exists}, \citecoq{forall}, \citecoq{->}, \citecoq{<-}, \texttt{\char'134/}, \texttt{/\char'134}, and \citecoq{=>}, respectively. For instance, the \coq{} statement %V8 A prendre % inclusion numero 1 % traduction numero 1 \begin{alltt} \hide{Open Scope nat_scope. Check (}forall A:Set,(exists x : A, forall (y:A), x <> y) -> 2 = 3\hide{).} \end{alltt} is written as follows in this tutorial: %V8 A prendre % inclusion numero 2 % traduction numero 2 \begin{alltt} \hide{Check (}{\prodsym}A:Set,(\exsym{}x:A, {\prodsym}y:A, x {\coqdiff} y) \arrow{} 2 = 3\hide{).} \end{alltt} When a fragment of \coq{} input text appears in the middle of regular text, we often place this fragment between double quotes ``\dots.'' These double quotes do not belong to the \coq{} syntax. Finally, any string enclosed between \texttt{(*} and \texttt{*)} is a comment and is ignored by the \coq{} system. \section{Introducing Inductive Types} \label{Introduction} Inductive types are types closed with respect to their introduction rules. These rules explain the most basic or \textsl{canonical} ways of constructing an element of the type. In this sense, they characterize the recursive type. Different rules must be considered as introducing different objects. In order to fix ideas, let us introduce in {\coq} the most well-known example of a recursive type: the type of natural numbers. %V8 A prendre \begin{alltt} Inductive nat : Set := | O : nat | S : nat\arrow{}nat. \end{alltt} The definition of a recursive type has two main parts. First, we establish what kind of recursive type we will characterize (a set, in this case). Second, we present the introduction rules that define the type ({\Z} and {\SUCC}), also called its {\sl constructors}. The constructors {\Z} and {\SUCC} determine all the elements of this type. In other words, if $n\mbox{:}\nat$, then $n$ must have been introduced either by the rule {\Z} or by an application of the rule {\SUCC} to a previously constructed natural number. In this sense, we can say that {\nat} is \emph{closed}. On the contrary, the type $\Set$ is an {\it open} type, since we do not know {\it a priori} all the possible ways of introducing an object of type \texttt{Set}. After entering this command, the constants {\nat}, {\Z} and {\SUCC} are available in the current context. We can see their types using the \texttt{Check} command \refmancite{Section \ref{Check}}: %V8 A prendre \begin{alltt} Check nat. \it{}nat : Set \tt{}Check O. \it{}O : nat \tt{}Check S. \it{}S : nat {\arrow} nat \end{alltt} Moreover, {\coq} adds to the context three constants named $\natind$, $\natrec$ and $\natrect$, which correspond to different principles of structural induction on natural numbers that {\coq} infers automatically from the definition. We will come back to them in Section \ref{StructuralInduction}. In fact, the type of natural numbers as well as several useful theorems about them are already defined in the basic library of {\coq}, so there is no need to introduce them. Therefore, let us throw away our (re)definition of {\nat}, using the command \texttt{Reset}. %V8 A prendre \begin{alltt} Reset nat. Print nat. \it{}Inductive nat : Set := O : nat | S : nat \arrow{} nat For S: Argument scope is [nat_scope] \end{alltt} Notice that \coq{}'s \emph{interpretation scope} for natural numbers (called \texttt{nat\_scope}) allows us to read and write natural numbers in decimal form (see \cite{coqrefman}). For instance, the constructor \texttt{O} can be read or written as the digit $0$, and the term ``~\texttt{S (S (S O))}~'' as $3$. %V8 A prendre \begin{alltt} Check O. \it 0 : nat. \tt Check (S (S (S O))). \it 3 : nat \end{alltt} Let us now take a look to some other recursive types contained in the standard library of {\coq}. \subsection{Lists} Lists are defined in library \citecoq{List}: \begin{alltt} Require Import List. Print list. \it Inductive list (A : Set) : Set := nil : list A | cons : A {\arrow} list A {\arrow} list A For nil: Argument A is implicit For cons: Argument A is implicit For list: Argument scope is [type_scope] For nil: Argument scope is [type_scope] For cons: Argument scopes are [type_scope _ _] \end{alltt} In this definition, \citecoq{A} is a \emph{general parameter}, global to both constructors. This kind of definition allows us to build a whole family of inductive types, indexed over the sort \citecoq{Set}. This can be observed if we consider the type of identifiers \citecoq{list}, \citecoq{cons} and \citecoq{nil}. Notice the notation \citecoq{(A := \dots)} which must be used when {\coq}'s type inference algorithm cannot infer the implicit parameter \citecoq{A}. \begin{alltt} Check list. \it list : Set {\arrow} Set \tt Check (nil (A:=nat)). \it nil : list nat \tt Check (nil (A:= nat {\arrow} nat)). \it nil : list (nat {\arrow} nat) \tt Check (fun A: Set {\funarrow} (cons (A:=A))). \it fun A : Set {\funarrow} cons (A:=A) : {\prodsym} A : Set, A {\arrow} list A {\arrow} list A \tt Check (cons 3 (cons 2 nil)). \it 3 :: 2 :: nil : list nat \end{alltt} \subsection{Vectors.} \label{vectors} Like \texttt{list}, \citecoq{vector} is a polymorphic type: if $A$ is a set, and $n$ a natural number, ``~\citecoq{vector $A$ $n$}~'' is the type of vectors of elements of $A$ and size $n$. \begin{alltt} Require Import Bvector. Print vector. \it Inductive vector (A : Set) : nat {\arrow} Set := Vnil : vector A 0 | Vcons : A {\arrow} {\prodsym} n : nat, vector A n {\arrow} vector A (S n) For vector: Argument scopes are [type_scope nat_scope] For Vnil: Argument scope is [type_scope] For Vcons: Argument scopes are [type_scope _ nat_scope _] \end{alltt} Remark the difference between the two parameters $A$ and $n$: The first one is a \textsl{general parameter}, global to all the introduction rules,while the second one is an \textsl{index}, which is instantiated differently in the introduction rules. Such types parameterized by regular values are called \emph{dependent types}. \begin{alltt} Check (Vnil nat). \it Vnil nat : vector nat 0 \tt Check (fun (A:Set)(a:A){\funarrow} Vcons _ a _ (Vnil _)). \it fun (A : Set) (a : A) {\funarrow} Vcons A a 0 (Vnil A) : {\prodsym} A : Set, A {\arrow} vector A 1 \tt Check (Vcons _ 5 _ (Vcons _ 3 _ (Vnil _))). \it Vcons nat 5 1 (Vcons nat 3 0 (Vnil nat)) : vector nat 2 \end{alltt} \subsection{The contradictory proposition.} Another example of an inductive type is the contradictory proposition. This type inhabits the universe of propositions, and has no element at all. %V8 A prendre \begin{alltt} Print False. \it{} Inductive False : Prop := \end{alltt} \noindent Notice that no constructor is given in this definition. \subsection{The tautological proposition.} Similarly, the tautological proposition {\True} is defined as an inductive type with only one element {\I}: %V8 A prendre \begin{alltt} Print True. \it{}Inductive True : Prop := I : True \end{alltt} \subsection{Relations as inductive types.} Some relations can also be introduced in a smart way as an inductive family of propositions. Let us take as example the order $n \leq m$ on natural numbers, called \citecoq{le} in {\coq}. This relation is introduced through the following definition, quoted from the standard library\footnote{In the interpretation scope for Peano arithmetic: \citecoq{nat\_scope}, ``~\citecoq{n <= m}~'' is equivalent to ``~\citecoq{le n m}~'' .}: %V8 A prendre \begin{alltt} Print le. \it Inductive le (n:nat) : nat\arrow{}Prop := | le_n: n {\coqle} n | le_S: {\prodsym} m, n {\coqle} m \arrow{} n {\coqle} S m. \end{alltt} Notice that in this definition $n$ is a general parameter, while the second argument of \citecoq{le} is an index (see section ~\ref{vectors}). This definition introduces the binary relation $n {\leq} m$ as the family of unary predicates ``\textsl{to be greater or equal than a given $n$}'', parameterized by $n$. The introduction rules of this type can be seen as a sort of Prolog rules for proving that a given integer $n$ is less or equal than another one. In fact, an object of type $n{\leq} m$ is nothing but a proof built up using the constructors \textsl{le\_n} and \textsl{le\_S} of this type. As an example, let us construct a proof that zero is less or equal than three using {\coq}'s interactive proof mode. Such an object can be obtained applying three times the second introduction rule of \citecoq{le}, to a proof that zero is less or equal than itself, which is provided by the first constructor of \citecoq{le}: %V8 A prendre \begin{alltt} Theorem zero_leq_three: 0 {\coqle} 3. Proof. \it{} 1 subgoal ============================ 0 {\coqle} 3 \tt{}Proof. constructor 2. \it{} 1 subgoal ============================ 0 {\coqle} 2 \tt{} constructor 2. \it{} 1 subgoal ============================ 0 {\coqle} 1 \tt{} constructor 2 \it{} 1 subgoal ============================ 0 {\coqle} 0 \tt{} constructor 1. \it{}Proof completed \tt{}Qed. \end{alltt} \noindent When the current goal is an inductive type, the tactic ``~\citecoq{constructor $i$}~'' \refmancite{Section \ref{constructor}} applies the $i$-th constructor in the definition of the type. We can take a look at the proof constructed using the command \texttt{Print}: %V8 A prendre \begin{alltt} Print Print zero_leq_three. \it{}zero_leq_three = zero_leq_three = le_S 0 2 (le_S 0 1 (le_S 0 0 (le_n 0))) : 0 {\coqle} 3 \end{alltt} When the parameter $i$ is not supplied, the tactic \texttt{constructor} tries to apply ``~\texttt{constructor $1$}~'', ``~\texttt{constructor $2$}~'',\dots, ``~\texttt{constructor $n$}~'' where $n$ is the number of constructors of the inductive type (2 in our example) of the conclusion of the goal. Our little proof can thus be obtained iterating the tactic \texttt{constructor} until it fails: %V8 A prendre \begin{alltt} Lemma zero_leq_three': 0 {\coqle} 3. repeat constructor. Qed. \end{alltt} Notice that the strict order on \texttt{nat}, called \citecoq{lt} is not inductively defined: \begin{alltt} Print lt. \it lt = fun n m : nat {\funarrow} S n {\coqle} m : nat {\arrow} nat {\arrow} Prop \tt Lemma zero_lt_three : 0 < 3. Proof. unfold lt. \it ==================== 1 {\coqle} 3 \tt repeat constructor. Qed. \end{alltt} \subsection{The propositional equality type.} \label{equality} In {\coq}, the propositional equality between two inhabitants $a$ and $b$ of the same type $A$ , noted $a=b$, is introduced as a family of recursive predicates ``~\textsl{to be equal to $a$}~'', parameterised by both $a$ and its type $A$. This family of types has only one introduction rule, which corresponds to reflexivity. Notice that the syntax ``\citecoq{$a$ = $b$}~'' is an abbreviation for ``\citecoq{eq $a$ $b$}~'', and that the parameter $A$ is \emph{implicit}, as it can be infered from $a$. %V8 A prendre \begin{alltt} Print eq. \it{} Inductive eq (A : Type) (x : A) : A \arrow{} Prop := refl_equal : x = x For eq: Argument A is implicit For refl_equal: Argument A is implicit For eq: Argument scopes are [type_scope _ _] For refl_equal: Argument scopes are [type_scope _] \end{alltt} Notice also that the first parameter $A$ of \texttt{eq} has type \texttt{Type}. The type system of {\coq} allows us to consider equality between various kinds of terms: elements of a set, proofs, propositions, types, and so on. Look at \cite{coqrefman, coqart} to get more details on {\coq}'s type system, as well as implicit arguments and argument scopes. \begin{alltt} Lemma eq_3_3 : 2 + 1 = 3. Proof. reflexivity. Qed. Lemma eq_proof_proof : refl_equal (2*6) = refl_equal (3*4). Proof. reflexivity. Qed. Print eq_proof_proof. \it eq_proof_proof = refl_equal (refl_equal (3 * 4)) : refl_equal (2 * 6) = refl_equal (3 * 4) \tt Lemma eq_lt_le : ( 2 < 4) = (3 {\coqle} 4). Proof. reflexivity. Qed. Lemma eq_nat_nat : nat = nat. Proof. reflexivity. Qed. Lemma eq_Set_Set : Set = Set. Proof. reflexivity. Qed. \end{alltt} \subsection{Logical connectives.} \label{LogicalConnectives} The conjunction and disjunction of two propositions are also examples of recursive types: \begin{alltt} Inductive or (A B : Prop) : Prop := or_introl : A \arrow{} A {\coqor} B | or_intror : B \arrow{} A {\coqor} B Inductive and (A B : Prop) : Prop := conj : A \arrow{} B \arrow{} A {\coqand} B \end{alltt} The propositions $A$ and $B$ are general parameters of these connectives. Choosing different universes for $A$ and $B$ and for the inductive type itself gives rise to different type constructors. For example, the type \textsl{sumbool} is a disjunction but with computational contents. \begin{alltt} Inductive sumbool (A B : Prop) : Set := left : A \arrow{} \{A\} + \{B\} | right : B \arrow{} \{A\} + \{B\} \end{alltt} This type --noted \texttt{\{$A$\}+\{$B$\}} in {\coq}-- can be used in {\coq} programs as a sort of boolean type, to check whether it is $A$ or $B$ that is true. The values ``~\citecoq{left $p$}~'' and ``~\citecoq{right $q$}~'' replace the boolean values \textsl{true} and \textsl{false}, respectively. The advantage of this type over \textsl{bool} is that it makes available the proofs $p$ of $A$ or $q$ of $B$, which could be necessary to construct a verification proof about the program. For instance, let us consider the certified program \citecoq{le\_lt\_dec} of the Standard Library. \begin{alltt} Require Import Compare_dec. Check le_lt_dec. \it le_lt_dec : {\prodsym} n m : nat, \{n {\coqle} m\} + \{m < n\} \end{alltt} We use \citecoq{le\_lt\_dec} to build a function for computing the max of two natural numbers: \begin{alltt} Definition max (n p :nat) := match le_lt_dec n p with | left _ {\funarrow} p | right _ {\funarrow} n end. \end{alltt} In the following proof, the case analysis on the term ``~\citecoq{le\_lt\_dec n p}~'' gives us an access to proofs of $n\leq p$ in the first case, $p % Cases p of % lt_intro1 {\funarrow} (lt_intro1 (S n)) % | (lt_intro2 m1 p2) {\funarrow} (lt_intro2 (S n) (S m1) (lt_n_S n m1 p2)) % end. %\end{alltt} %The guardedness condition must be satisfied only by the last argument %of the enclosed list. For example, the following declaration is an %alternative way of defining addition: %\begin{alltt} %Reset add. %Fixpoint add [n:nat] : nat\arrow{}nat := % Cases n of % O {\funarrow} [x:nat]x % | (S m) {\funarrow} [x:nat](add m (S x)) % end. %\end{alltt} In the following definition of addition, the second argument of \verb@plus''@ grows at each recursive call. However, as the first one always decreases, the definition is sound. \begin{alltt} Fixpoint plus'' (n p:nat) \{struct n\} : nat := match n with | 0 {\funarrow} p | S m {\funarrow} plus'' m (S p) end. \end{alltt} Moreover, the argument in the recursive call could be a deeper component of $n$. This is the case in the following definition of a boolean function determining whether a number is even or odd: \begin{alltt} Fixpoint even_test (n:nat) : bool := match n with 0 {\funarrow} true | 1 {\funarrow} false | S (S p) {\funarrow} even_test p end. \end{alltt} Mutually dependent definitions by structural induction are also allowed. For example, the previous function \textsl{even} could alternatively be defined using an auxiliary function \textsl{odd}: \begin{alltt} Reset even_test. Fixpoint even_test (n:nat) : bool := match n with | 0 {\funarrow} true | S p {\funarrow} odd_test p end with odd_test (n:nat) : bool := match n with | 0 {\funarrow} false | S p {\funarrow} even_test p end. \end{alltt} %\begin{exercise} %Define a function by structural induction that computes the number of %nodes of a tree structure defined in page \pageref{Forest}. %\end{exercise} Definitions by structural induction are computed only when they are applied, and the decreasing argument is a term having a constructor at the head. We can check this using the \texttt{Eval} command, which computes the normal form of a well typed term. \begin{alltt} Eval simpl in even_test. \it = even_test : nat {\arrow} bool \tt Eval simpl in (fun x : nat {\funarrow} even x). \it = fun x : nat {\funarrow} even x : nat {\arrow} Prop \tt Eval simpl in (fun x : nat => plus 5 x). \it = fun x : nat {\funarrow} S (S (S (S (S x)))) \tt Eval simpl in (fun x : nat {\funarrow} even_test (plus 5 x)). \it = fun x : nat {\funarrow} odd_test x : nat {\arrow} bool \tt Eval simpl in (fun x : nat {\funarrow} even_test (plus x 5)). \it = fun x : nat {\funarrow} even_test (x + 5) : nat {\arrow} bool \end{alltt} %\begin{exercise} %Prove that the second definition of even satisfies the following %theorem: %\begin{verbatim} %Theorem unfold_even : % (x:nat) % (even x)= (Cases x of % O {\funarrow} true % | (S O) {\funarrow} false % | (S (S m)) {\funarrow} (even m) % end). %\end{verbatim} %\end{exercise} \subsection{Proofs by Structural Induction} The principle of structural induction can be also used in order to define proofs, that is, to prove theorems. Let us call an \textsl{elimination combinator} any function that, given a predicate $P$, defines a proof of ``~$P\;x$~'' by structural induction on $x$. In {\coq}, the principle of proof by induction on natural numbers is a particular case of an elimination combinator. The definition of this combinator depends on three general parameters: the predicate to be proven, the base case, and the inductive step: \begin{alltt} Section Principle_of_Induction. Variable P : nat {\arrow} Prop. Hypothesis base_case : P 0. Hypothesis inductive_step : {\prodsym} n:nat, P n {\arrow} P (S n). Fixpoint nat_ind (n:nat) : (P n) := match n return P n with | 0 {\funarrow} base_case | S m {\funarrow} inductive_step m (nat_ind m) end. End Principle_of_Induction. \end{alltt} As this proof principle is used very often, {\coq} automatically generates it when an inductive type is introduced. Similar principles \texttt{nat\_rec} and \texttt{nat\_rect} for defining objects in the universes $\Set$ and $\Type$ are also automatically generated \footnote{In fact, whenever possible, {\coq} generates the principle \texttt{$I$\_rect}, then derives from it the weaker principles \texttt{$I$\_ind} and \texttt{$I$\_rec}. If some principle has to be defined by hand, the user may try to build \texttt{$I$\_rect} (if possible). Thanks to {\coq}'s conversion rule, this principle can be used directly to build proofs and/or programs.}. The command \texttt{Scheme} \refmancite{Section \ref{Scheme}} can be used to generate an elimination combinator from certain parameters, like the universe that the defined objects must inhabit, whether the case analysis in the definitions must be dependent or not, etc. For example, it can be used to generate an elimination combinator for reasoning on even natural numbers from the mutually dependent predicates introduced in page \pageref{Even}. We do not display the combinators here by lack of space, but you can see them using the \texttt{Print} command. \begin{alltt} Scheme Even_induction := Minimality for even Sort Prop with Odd_induction := Minimality for odd Sort Prop. \end{alltt} \begin{alltt} Theorem even_plus_four : {\prodsym} n:nat, even n {\arrow} even (4+n). Proof. intros n H. elim H using Even_induction with (P0 := fun n {\funarrow} odd (4+n)); simpl;repeat constructor;assumption. Qed. \end{alltt} Another example of an elimination combinator is the principle of double induction on natural numbers, introduced by the following definition: \begin{alltt} Section Principle_of_Double_Induction. Variable P : nat {\arrow} nat {\arrow}Prop. Hypothesis base_case1 : {\prodsym} m:nat, P 0 m. Hypothesis base_case2 : {\prodsym} n:nat, P (S n) 0. Hypothesis inductive_step : {\prodsym} n m:nat, P n m {\arrow} \,\, P (S n) (S m). Fixpoint nat_double_ind (n m:nat)\{struct n\} : P n m := match n, m return P n m with | 0 , x {\funarrow} base_case1 x | (S x), 0 {\funarrow} base_case2 x | (S x), (S y) {\funarrow} inductive_step x y (nat_double_ind x y) end. End Principle_of_Double_Induction. \end{alltt} Changing the type of $P$ into $\nat\rightarrow\nat\rightarrow\Set$, another combinator \texttt{nat\_double\_rec} for constructing (certified) programs can be defined in exactly the same way. This definition is left as an exercise.\label{natdoublerec} \iffalse \begin{alltt} Section Principle_of_Double_Recursion. Variable P : nat {\arrow} nat {\arrow} Set. Hypothesis base_case1 : {\prodsym} x:nat, P 0 x. Hypothesis base_case2 : {\prodsym} x:nat, P (S x) 0. Hypothesis inductive_step : {\prodsym} n m:nat, P n m {\arrow} P (S n) (S m). Fixpoint nat_double_rec (n m:nat)\{struct n\} : P n m := match n, m return P n m with 0 , x {\funarrow} base_case1 x | (S x), 0 {\funarrow} base_case2 x | (S x), (S y) {\funarrow} inductive_step x y (nat_double_rec x y) end. End Principle_of_Double_Recursion. \end{alltt} \fi For instance the function computing the minimum of two natural numbers can be defined in the following way: \begin{alltt} Definition min : nat {\arrow} nat {\arrow} nat := nat_double_rec (fun (x y:nat) {\funarrow} nat) (fun (x:nat) {\funarrow} 0) (fun (y:nat) {\funarrow} 0) (fun (x y r:nat) {\funarrow} S r). Eval compute in (min 5 8). \it = 5 : nat \end{alltt} %\begin{exercise} % %Define the combinator \texttt{nat\_double\_rec}, and apply it %to give another definition of \citecoq{le\_lt\_dec} (using the theorems %of the \texttt{Arith} library). %\end{exercise} \subsection{Using Elimination Combinators.} The tactic \texttt{apply} can be used to apply one of these proof principles during the development of a proof. \begin{alltt} Lemma not_circular : {\prodsym} n:nat, n {\coqdiff} S n. Proof. intro n. apply nat_ind with (P:= fun n {\funarrow} n {\coqdiff} S n). \it 2 subgoals n : nat ============================ 0 {\coqdiff} 1 subgoal 2 is: {\prodsym} n0 : nat, n0 {\coqdiff} S n0 {\arrow} S n0 {\coqdiff} S (S n0) \tt discriminate. red; intros n0 Hn0 eqn0Sn0;injection eqn0Sn0;trivial. Qed. \end{alltt} The tactic \texttt{elim} \refmancite{Section \ref{Elim}} is a refinement of \texttt{apply}, specially designed for the application of elimination combinators. If $t$ is an object of an inductive type $I$, then ``~\citecoq{elim $t$}~'' tries to find an abstraction $P$ of the current goal $G$ such that $(P\;t)\equiv G$. Then it solves the goal applying ``~$I\texttt{\_ind}\;P$~'', where $I$\texttt{\_ind} is the combinator associated to $I$. The different cases of the induction then appear as subgoals that remain to be solved. In the previous proof, the tactic call ``~\citecoq{apply nat\_ind with (P:= fun n {\funarrow} n {\coqdiff} S n)}~'' can simply be replaced with ``~\citecoq{elim n}~''. The option ``~\citecoq{\texttt{elim} $t$ \texttt{using} $C$}~'' allows to use a derived combinator $C$ instead of the default one. Consider the following theorem, stating that equality is decidable on natural numbers: \label{iseqpage} \begin{alltt} Lemma eq_nat_dec : {\prodsym} n p:nat, \{n=p\}+\{n {\coqdiff} p\}. Proof. intros n p. \end{alltt} Let us prove this theorem using the combinator \texttt{nat\_double\_rec} of section~\ref{natdoublerec}. The example also illustrates how \texttt{elim} may sometimes fail in finding a suitable abstraction $P$ of the goal. Note that if ``~\texttt{elim n}~'' is used directly on the goal, the result is not the expected one. \vspace{12pt} %\pagebreak \begin{alltt} elim n using nat_double_rec. \it 4 subgoals n : nat p : nat ============================ {\prodsym} x : nat, \{x = p\} + \{x {\coqdiff} p\} subgoal 2 is: nat {\arrow} \{0 = p\} + \{0 {\coqdiff} p\} subgoal 3 is: nat {\arrow} {\prodsym} m : nat, \{m = p\} + \{m {\coqdiff} p\} {\arrow} \{S m = p\} + \{S m {\coqdiff} p\} subgoal 4 is: nat \end{alltt} The four sub-goals obtained do not correspond to the premises that would be expected for the principle \texttt{nat\_double\_rec}. The problem comes from the fact that this principle for eliminating $n$ has a universally quantified formula as conclusion, which confuses \texttt{elim} about the right way of abstracting the goal. %In effect, let us consider the type of the goal before the call to %\citecoq{elim}: ``~\citecoq{\{n = p\} + \{n {\coqdiff} p\}}~''. %Among all the abstractions that can be built by ``~\citecoq{elim n}~'' %let us consider this one %$P=$\citecoq{fun n :nat {\funarrow} fun q : nat {\funarrow} {\{q= p\} + \{q {\coqdiff} p\}}}. %It is easy to verify that %$P$ has type \citecoq{nat {\arrow} nat {\arrow} Set}, and that, if some %$q:\citecoq{nat}$ is given, then $P\;q\;$ matches the current goal. %Then applying \citecoq{nat\_double\_rec} with $P$ generates %four goals, corresponding to Therefore, in this case the abstraction must be explicited using the tactic \texttt{pattern}. Once the right abstraction is provided, the rest of the proof is immediate: \begin{alltt} Undo. pattern p,n. \it n : nat p : nat ============================ (fun n0 n1 : nat {\funarrow} \{n1 = n0\} + \{n1 {\coqdiff} n0\}) p n \tt elim n using nat_double_rec. \it 3 subgoals n : nat p : nat ============================ {\prodsym} x : nat, \{x = 0\} + \{x {\coqdiff} 0\} subgoal 2 is: {\prodsym} x : nat, \{0 = S x\} + \{0 {\coqdiff} S x\} subgoal 3 is: {\prodsym} n0 m : nat, \{m = n0\} + \{m {\coqdiff} n0\} {\arrow} \{S m = S n0\} + \{S m {\coqdiff} S n0\} \tt destruct x; auto. destruct x; auto. intros n0 m H; case H. intro eq; rewrite eq ; auto. intro neg; right; red ; injection 1; auto. Defined. \end{alltt} Notice that the tactic ``~\texttt{decide equality}~'' \refmancite{Section\ref{DecideEquality}} generalises the proof above to a large class of inductive types. It can be used for proving a proposition of the form $\forall\,(x,y:R),\{x=y\}+\{x{\coqdiff}y\}$, where $R$ is an inductive datatype all whose constructors take informative arguments ---like for example the type {\nat}: \begin{alltt} Definition eq_nat_dec' : {\prodsym} n p:nat, \{n=p\} + \{n{\coqdiff}p\}. decide equality. Defined. \end{alltt} \begin{exercise} \begin{enumerate} \item Define a recursive function \emph{nat2itree} mapping any natural number $n$ into an infinitely branching tree of height $n$. \item Provide an elimination combinator for these trees. \item Prove that the relation \citecoq{itree\_le} is a preorder (i.e. reflexive and transitive). \end{enumerate} \end{exercise} \begin{exercise} \label{zeroton} Define the type of lists, and a predicate ``being an ordered list'' using an inductive family. Then, define the function $(from\;n)=0::1\;\ldots\; n::\texttt{nil}$ and prove that it always generates an ordered list. \end{exercise} \subsection{Well-founded Recursion} \label{WellFoundedRecursion} Structural induction is a strong elimination rule for inductive types. This method can be used to define any function whose termination is based on the well-foundedness of certain order relation $R$ decreasing at each recursive call. What makes this principle so strong is the possibility of reasoning by structural induction on the proof that certain $R$ is well-founded. In order to illustrate this we have first to introduce the predicate of accessibility. \begin{alltt} Print Acc. \it Inductive Acc (A : Set) (R : A {\arrow} A {\arrow} Prop) (x:A) : Prop := Acc_intro : ({\prodsym} y : A, R y x {\arrow} Acc R y) {\arrow} Acc R x For Acc: Argument A is implicit For Acc_intro: Arguments A, R are implicit \dots \end{alltt} \noindent This inductive predicate characterize those elements $x$ of $A$ such that any descending $R$-chain $\ldots x_2\;R\;x_1\;R\;x$ starting from $x$ is finite. A well-founded relation is a relation such that all the elements of $A$ are accessible. Consider now the problem of representing in {\coq} the following ML function $\textsl{div}(x,y)$ on natural numbers, which computes $\lceil\frac{x}{y}\rceil$ if $y>0$ and yields $x$ otherwise. \begin{verbatim} let rec div x y = if x = 0 then 0 else if y = 0 then x else (div (x-y) y)+1;; \end{verbatim} The equality test on natural numbers can be represented as the function \textsl{eq\_nat\_dec} defined page \pageref{iseqpage}. Giving $x$ and $y$, this function yields either the value $(\textsl{left}\;p)$ if there exists a proof $p:x=y$, or the value $(\textsl{right}\;q)$ if there exists $q:a\not = b$. The subtraction function is already defined in the library \citecoq{Minus}. Hence, direct translation of the ML function \textsl{div} would be: \begin{alltt} Require Import Minus. Fixpoint div (x y:nat)\{struct x\}: nat := if eq_nat_dec x 0 then 0 else if eq_nat_dec y 0 then x else S (div (x-y) y). \it Error: Recursive definition of div is ill-formed. In environment div : nat {\arrow} nat {\arrow} nat x : nat y : nat _ : x {\coqdiff} 0 _ : y {\coqdiff} 0 Recursive call to div has principal argument equal to "x - y" instead of a subterm of x \end{alltt} The program \texttt{div} is rejected by {\coq} because it does not verify the syntactical condition to ensure termination. In particular, the argument of the recursive call is not a pattern variable issued from a case analysis on $x$. We would have the same problem if we had the directive ``~\citecoq{\{struct y\}}~'' instead of ``~\citecoq{\{struct x\}}~''. However, we know that this program always stops. One way to justify its termination is to define it by structural induction on a proof that $x$ is accessible trough the relation $<$. Notice that any natural number $x$ is accessible for this relation. In order to do this, it is first necessary to prove some auxiliary lemmas, justifying that the first argument of \texttt{div} decreases at each recursive call. \begin{alltt} Lemma minus_smaller_S : {\prodsym} x y:nat, x - y < S x. Proof. intros x y; pattern y, x; elim x using nat_double_ind. destruct x0; auto with arith. simpl; auto with arith. simpl; auto with arith. Qed. Lemma minus_smaller_positive : {\prodsym} x y:nat, x {\coqdiff}0 {\arrow} y {\coqdiff} 0 {\arrow} x - y < x. Proof. destruct x; destruct y; ( simpl;intros; apply minus_smaller || intros; absurd (0=0); auto). Qed. \end{alltt} \noindent The last two lemmas are necessary to prove that for any pair of positive natural numbers $x$ and $y$, if $x$ is accessible with respect to \citecoq{lt}, then so is $x-y$. \begin{alltt} Definition minus_decrease : {\prodsym} x y:nat, Acc lt x {\arrow} x {\coqdiff} 0 {\arrow} y {\coqdiff} 0 {\arrow} Acc lt (x-y). Proof. intros x y H; case H. intros Hz posz posy. apply Hz; apply minus_smaller_positive; assumption. Defined. \end{alltt} Let us take a look at the proof of the lemma \textsl{minus\_decrease}, since the way in which it has been proven is crucial for what follows. \begin{alltt} Print minus_decrease. \it minus_decrease = fun (x y : nat) (H : Acc lt x) {\funarrow} match H in (Acc _ y0) return (y0 {\coqdiff} 0 {\arrow} y {\coqdiff} 0 {\arrow} Acc lt (y0 - y)) with | Acc_intro z Hz {\funarrow} fun (posz : z {\coqdiff} 0) (posy : y {\coqdiff} 0) {\funarrow} Hz (z - y) (minus_smaller_positive z y posz posy) end : {\prodsym} x y : nat, Acc lt x {\arrow} x {\coqdiff} 0 {\arrow} y {\coqdiff} 0 {\arrow} Acc lt (x - y) \end{alltt} \noindent Notice that the function call $(\texttt{minus\_decrease}\;n\;m\;H)$ indeed yields an accessibility proof that is \textsl{structurally smaller} than its argument $H$, because it is (an application of) its recursive component $Hz$. This enables to justify the following definition of \textsl{div\_aux}: \begin{alltt} Definition div_aux (x y:nat)(H: Acc lt x):nat. fix 3. intros. refine (if eq_nat_dec x 0 then 0 else if eq_nat_dec y 0 then y else div_aux (x-y) y _). \it div_aux : {\prodsym} x : nat, nat {\arrow} Acc lt x {\arrow} nat x : nat y : nat H : Acc lt x _ : x {\coqdiff} 0 _0 : y {\coqdiff} 0 ============================ Acc lt (x - y) \tt apply (minus_decrease x y H);auto. Defined. \end{alltt} The main division function is easily defined, using the theorem \citecoq{lt\_wf} of the library \citecoq{Wf\_nat}. This theorem asserts that \citecoq{nat} is well founded w.r.t. \citecoq{lt}, thus any natural number is accessible. \begin{alltt} Definition div x y := div_aux x y (lt_wf x). \end{alltt} Let us explain the proof above. In the definition of \citecoq{div\_aux}, what decreases is not $x$ but the \textsl{proof} of the accessibility of $x$. The tactic ``~\texttt{fix 3}~'' is used to indicate that the proof proceeds by structural induction on the third argument of the theorem --that is, on the accessibility proof. It also introduces a new hypothesis in the context, named as the current theorem, and with the same type as the goal. Then, the proof is refined with an incomplete proof term, containing a hole \texttt{\_}. This hole corresponds to the proof of accessibility for $x-y$, and is filled up with the (smaller!) accessibility proof provided by the function \texttt{minus\_decrease}. \noindent Let us take a look to the term \textsl{div\_aux} defined: \pagebreak \begin{alltt} Print div_aux. \it div_aux = (fix div_aux (x y : nat) (H : Acc lt x) \{struct H\} : nat := match eq_nat_dec x 0 with | left _ {\funarrow} 0 | right _ {\funarrow} match eq_nat_dec y 0 with | left _ {\funarrow} y | right _0 {\funarrow} div_aux (x - y) y (minus_decrease x y H _ _0) end end) : {\prodsym} x : nat, nat {\arrow} Acc lt x {\arrow} nat \end{alltt} If the non-informative parts from this proof --that is, the accessibility proof-- are erased, then we obtain exactly the program that we were looking for. \begin{alltt} Extraction div. \it let div x y = div_aux x y \tt Extraction div_aux. \it let rec div_aux x y = match eq_nat_dec x O with | Left {\arrow} O | Right {\arrow} (match eq_nat_dec y O with | Left {\arrow} y | Right {\arrow} div_aux (minus x y) y) \end{alltt} This methodology enables the representation of any program whose termination can be proved in {\coq}. Once the expected properties from this program have been verified, the justification of its termination can be thrown away, keeping just the desired computational behavior for it. \section{A case study in dependent elimination}\label{CaseStudy} Dependent types are very expressive, but ignoring some useful techniques can cause some problems to the beginner. Let us consider again the type of vectors (see section~\ref{vectors}). We want to prove a quite trivial property: the only value of type ``~\citecoq{vector A 0}~'' is ``~\citecoq{Vnil $A$}~''. Our first naive attempt leads to a \emph{cul-de-sac}. \begin{alltt} Lemma vector0_is_vnil : {\prodsym} (A:Set)(v:vector A 0), v = Vnil A. Proof. intros A v;inversion v. \it 1 subgoal A : Set v : vector A 0 ============================ v = Vnil A \tt Abort. \end{alltt} Another attempt is to do a case analysis on a vector of any length $n$, under an explicit hypothesis $n=0$. The tactic \texttt{discriminate} will help us to get rid of the case $n=\texttt{S $p$}$. Unfortunately, even the statement of our lemma is refused! \begin{alltt} Lemma vector0_is_vnil_aux : {\prodsym} (A:Set)(n:nat)(v:vector A n), n = 0 {\arrow} v = Vnil A. \it Error: In environment A : Set n : nat v : vector A n e : n = 0 The term "Vnil A" has type "vector A 0" while it is expected to have type "vector A n" \end{alltt} In effect, the equality ``~\citecoq{v = Vnil A}~'' is ill typed, because the type ``~\citecoq{vector A n}~'' is not \emph{convertible} with ``~\citecoq{vector A 0}~''. This problem can be solved if we consider the heterogeneous equality \citecoq{JMeq} \cite{conor:motive} which allows us to consider terms of different types, even if this equality can only be proven for terms in the same type. The axiom \citecoq{JMeq\_eq}, from the library \citecoq{JMeq} allows us to convert a heterogeneous equality to a standard one. \begin{alltt} Lemma vector0_is_vnil_aux : {\prodsym} (A:Set)(n:nat)(v:vector A n), n= 0 {\arrow} JMeq v (Vnil A). Proof. destruct v. auto. intro; discriminate. Qed. \end{alltt} Our property of vectors of null length can be easily proven: \begin{alltt} Lemma vector0_is_vnil : {\prodsym} (A:Set)(v:vector A 0), v = Vnil A. intros a v;apply JMeq_eq. apply vector0_is_vnil_aux. trivial. Qed. \end{alltt} It is interesting to look at another proof of \citecoq{vector0\_is\_vnil}, which illustrates a technique developed and used by various people (consult in the \emph{Coq-club} mailing list archive the contributions by Yves Bertot, Pierre Letouzey, Laurent Théry, Jean Duprat, and Nicolas Magaud, Venanzio Capretta and Conor McBride). This technique is also used for unfolding infinite list definitions (see chapter13 of~\cite{coqart}). Notice that this definition does not rely on any axiom (\emph{e.g.} \texttt{JMeq\_eq}). We first give a new definition of the identity on vectors. Before that, we make the use of constructors and selectors lighter thanks to the implicit arguments feature: \begin{alltt} Implicit Arguments Vcons [A n]. Implicit Arguments Vnil [A]. Implicit Arguments Vhead [A n]. Implicit Arguments Vtail [A n]. Definition Vid : {\prodsym} (A : Set)(n:nat), vector A n {\arrow} vector A n. Proof. destruct n; intro v. exact Vnil. exact (Vcons (Vhead v) (Vtail v)). Defined. \end{alltt} Then we prove that \citecoq{Vid} is the identity on vectors: \begin{alltt} Lemma Vid_eq : {\prodsym} (n:nat) (A:Set)(v:vector A n), v=(Vid _ n v). Proof. destruct v. \it A : Set ============================ Vnil = Vid A 0 Vnil subgoal 2 is: Vcons a v = Vid A (S n) (Vcons a v) \tt reflexivity. reflexivity. Defined. \end{alltt} Why defining a new identity function on vectors? The following dialogue shows that \citecoq{Vid} has some interesting computational properties: \begin{alltt} Eval simpl in (fun (A:Set)(v:vector A 0) {\funarrow} (Vid _ _ v)). \it = fun (A : Set) (_ : vector A 0) {\funarrow} Vnil : {\prodsym} A : Set, vector A 0 {\arrow} vector A 0 \end{alltt} Notice that the plain identity on vectors doesn't convert \citecoq{v} into \citecoq{Vnil}. \begin{alltt} Eval simpl in (fun (A:Set)(v:vector A 0) {\funarrow} v). \it = fun (A : Set) (v : vector A 0) {\funarrow} v : {\prodsym} A : Set, vector A 0 {\arrow} vector A 0 \end{alltt} Then we prove easily that any vector of length 0 is \citecoq{Vnil}: \begin{alltt} Theorem zero_nil : {\prodsym} A (v:vector A 0), v = Vnil. Proof. intros. change (Vnil (A:=A)) with (Vid _ 0 v). \it 1 subgoal A : Set v : vector A 0 ============================ v = Vid A 0 v \tt apply Vid_eq. Defined. \end{alltt} A similar result can be proven about vectors of strictly positive lenght\footnote{As for \citecoq{Vid} and \citecoq{Vid\_eq}, this definition is from Jean Duprat.}. \begin{alltt} Theorem decomp : {\prodsym} (A : Set) (n : nat) (v : vector A (S n)), v = Vcons (Vhead v) (Vtail v). Proof. intros. change (Vcons (Vhead v) (Vtail v)) with (Vid _ (S n) v). \it 1 subgoal A : Set n : nat v : vector A (S n) ============================ v = Vid A (S n) v \tt{} apply Vid_eq. Defined. \end{alltt} Both lemmas: \citecoq{zero\_nil} and \citecoq{decomp}, can be used to easily derive a double recursion principle on vectors of same length: \begin{alltt} Definition vector_double_rect : {\prodsym} (A:Set) (P: {\prodsym} (n:nat),(vector A n){\arrow}(vector A n) {\arrow} Type), P 0 Vnil Vnil {\arrow} ({\prodsym} n (v1 v2 : vector A n) a b, P n v1 v2 {\arrow} P (S n) (Vcons a v1) (Vcons b v2)) {\arrow} {\prodsym} n (v1 v2 : vector A n), P n v1 v2. induction n. intros; rewrite (zero_nil _ v1); rewrite (zero_nil _ v2). auto. intros v1 v2; rewrite (decomp _ _ v1);rewrite (decomp _ _ v2). apply X0; auto. Defined. \end{alltt} Notice that, due to the conversion rule of {\coq}'s type system, this function can be used directly with \citecoq{Prop} or \citecoq{Set} instead of type (thus it is useless to build \citecoq{vector\_double\_ind} and \citecoq{vector\_double\_rec}) from scratch. We finish this example with showing how to define the bitwise \emph{or} on boolean vectors of the same length, and proving a little property about this operation. \begin{alltt} Definition bitwise_or n v1 v2 : vector bool n := vector_double_rect bool (fun n v1 v2 {\funarrow} vector bool n) Vnil (fun n v1 v2 a b r {\funarrow} Vcons (orb a b) r) n v1 v2. \end{alltt} Let us define recursively the $n$-th element of a vector. Notice that it must be a partial function, in case $n$ is greater or equal than the length of the vector. Since {\coq} only considers total functions, the function returns a value in an \emph{option} type. \begin{alltt} Fixpoint vector_nth (A:Set)(n:nat)(p:nat)(v:vector A p) \{struct v\} : option A := match n,v with _ , Vnil {\funarrow} None | 0 , Vcons b _ _ {\funarrow} Some b | S n', Vcons _ p' v' {\funarrow} vector_nth A n' p' v' end. Implicit Arguments vector_nth [A p]. \end{alltt} We can now prove --- using the double induction combinator --- a simple property relying \citecoq{vector\_nth} and \citecoq{bitwise\_or}: \begin{alltt} Lemma nth_bitwise : {\prodsym} (n:nat) (v1 v2: vector bool n) i a b, vector_nth i v1 = Some a {\arrow} vector_nth i v2 = Some b {\arrow} vector_nth i (bitwise_or _ v1 v2) = Some (orb a b). Proof. intros n v1 v2; pattern n,v1,v2. apply vector_double_rect. simpl. destruct i; discriminate 1. destruct i; simpl;auto. injection 1; injection 2;intros; subst a; subst b; auto. Qed. \end{alltt} \section{Co-inductive Types and Non-ending Constructions} \label{CoInduction} The objects of an inductive type are well-founded with respect to the constructors of the type. In other words, these objects are built by applying \emph{a finite number of times} the constructors of the type. Co-inductive types are obtained by relaxing this condition, and may contain non-well-founded objects \cite{EG96,EG95a}. An example of a co-inductive type is the type of infinite sequences formed with elements of type $A$, also called streams. This type can be introduced through the following definition: \begin{alltt} CoInductive Stream (A: Set) :Set := | Cons : A\arrow{}Stream A\arrow{}Stream A. \end{alltt} If we are interested in finite or infinite sequences, we consider the type of \emph{lazy lists}: \begin{alltt} CoInductive LList (A: Set) : Set := | LNil : LList A | LCons : A {\arrow} LList A {\arrow} LList A. \end{alltt} It is also possible to define co-inductive types for the trees with infinite branches (see Chapter 13 of~\cite{coqart}). Structural induction is the way of expressing that inductive types only contain well-founded objects. Hence, this elimination principle is not valid for co-inductive types, and the only elimination rule for streams is case analysis. This principle can be used, for example, to define the destructors \textsl{head} and \textsl{tail}. \begin{alltt} Definition head (A:Set)(s : Stream A) := match s with Cons a s' {\funarrow} a end. Definition tail (A : Set)(s : Stream A) := match s with Cons a s' {\funarrow} s' end. \end{alltt} Infinite objects are defined by means of (non-ending) methods of construction, like in lazy functional programming languages. Such methods can be defined using the \texttt{CoFixpoint} command \refmancite{Section \ref{CoFixpoint}}. For example, the following definition introduces the infinite list $[a,a,a,\ldots]$: \begin{alltt} CoFixpoint repeat (A:Set)(a:A) : Stream A := Cons a (repeat a). \end{alltt} However, not every co-recursive definition is an admissible method of construction. Similarly to the case of structural induction, the definition must verify a \textsl{guardedness} condition to be accepted. This condition states that any recursive call in the definition must be protected --i.e, be an argument of-- some constructor, and only an argument of constructors \cite{EG94a}. The following definitions are examples of valid methods of construction: \begin{alltt} CoFixpoint iterate (A: Set)(f: A {\arrow} A)(a : A) : Stream A:= Cons a (iterate f (f a)). CoFixpoint map (A B:Set)(f: A {\arrow} B)(s : Stream A) : Stream B:= match s with Cons a tl {\funarrow} Cons (f a) (map f tl) end. \end{alltt} \begin{exercise} Define two different methods for constructing the stream which infinitely alternates the values \citecoq{true} and \citecoq{false}. \end{exercise} \begin{exercise} Using the destructors \texttt{head} and \texttt{tail}, define a function which takes the n-th element of an infinite stream. \end{exercise} A non-ending method of construction is computed lazily. This means that its definition is unfolded only when the object that it introduces is eliminated, that is, when it appears as the argument of a case expression. We can check this using the command \texttt{Eval}. \begin{alltt} Eval simpl in (fun (A:Set)(a:A) {\funarrow} repeat a). \it = fun (A : Set) (a : A) {\funarrow} repeat a : {\prodsym} A : Set, A {\arrow} Stream A \tt Eval simpl in (fun (A:Set)(a:A) {\funarrow} head (repeat a)). \it = fun (A : Set) (a : A) {\funarrow} a : {\prodsym} A : Set, A {\arrow} A \end{alltt} %\begin{exercise} %Prove the following theorem: %\begin{verbatim} %Theorem expand_repeat : (a:A)(repeat a)=(Cons a (repeat a)). %\end{verbatim} %Hint: Prove first the streams version of the lemma in exercise %\ref{expand}. %\end{exercise} \subsection{Extensional Properties} Case analysis is also a valid proof principle for infinite objects. However, this principle is not sufficient to prove \textsl{extensional} properties, that is, properties concerning the whole infinite object \cite{EG95a}. A typical example of an extensional property is the predicate expressing that two streams have the same elements. In many cases, the minimal reflexive relation $a=b$ that is used as equality for inductive types is too small to capture equality between streams. Consider for example the streams $\texttt{iterate}\;f\;(f\;x)$ and $(\texttt{map}\;f\;(\texttt{iterate}\;f\;x))$. Even though these two streams have the same elements, no finite expansion of their definitions lead to equal terms. In other words, in order to deal with extensional properties, it is necessary to construct infinite proofs. The type of infinite proofs of equality can be introduced as a co-inductive predicate, as follows: \begin{alltt} CoInductive EqSt (A: Set) : Stream A {\arrow} Stream A {\arrow} Prop := eqst : {\prodsym} s1 s2: Stream A, head s1 = head s2 {\arrow} EqSt (tail s1) (tail s2) {\arrow} EqSt s1 s2. \end{alltt} It is possible to introduce proof principles for reasoning about infinite objects as combinators defined through \texttt{CoFixpoint}. However, oppositely to the case of inductive types, proof principles associated to co-inductive types are not elimination but \textsl{introduction} combinators. An example of such a combinator is Park's principle for proving the equality of two streams, usually called the \textsl{principle of co-induction}. It states that two streams are equal if they satisfy a \textit{bisimulation}. A bisimulation is a binary relation $R$ such that any pair of streams $s_1$ ad $s_2$ satisfying $R$ have equal heads, and tails also satisfying $R$. This principle is in fact a method for constructing an infinite proof: \begin{alltt} Section Parks_Principle. Variable A : Set. Variable R : Stream A {\arrow} Stream A {\arrow} Prop. Hypothesis bisim1 : {\prodsym} s1 s2:Stream A, R s1 s2 {\arrow} head s1 = head s2. Hypothesis bisim2 : {\prodsym} s1 s2:Stream A, R s1 s2 {\arrow} R (tail s1) (tail s2). CoFixpoint park_ppl : {\prodsym} s1 s2:Stream A, R s1 s2 {\arrow} EqSt s1 s2 := fun s1 s2 (p : R s1 s2) {\funarrow} eqst s1 s2 (bisim1 s1 s2 p) (park_ppl (tail s1) (tail s2) (bisim2 s1 s2 p)). End Parks_Principle. \end{alltt} Let us use the principle of co-induction to prove the extensional equality mentioned above. \begin{alltt} Theorem map_iterate : {\prodsym} (a:Set)(f:A{\arrow}A)(x:A), EqSt (iterate f (f x)) (map f (iterate f x)). Proof. intros A f x. apply park_ppl with (R:= fun s1 s2 {\funarrow} {\exsym} x: A, s1 = iterate f (f x) {\coqand} s2 = map f (iterate f x)). intros s1 s2 (x0,(eqs1,eqs2)); rewrite eqs1; rewrite eqs2; reflexivity. intros s1 s2 (x0,(eqs1,eqs2)). exists (f x0);split; [rewrite eqs1|rewrite eqs2]; reflexivity. exists x;split; reflexivity. Qed. \end{alltt} The use of Park's principle is sometimes annoying, because it requires to find an invariant relation and prove that it is indeed a bisimulation. In many cases, a shorter proof can be obtained trying to construct an ad-hoc infinite proof, defined by a guarded declaration. The tactic ``~``\texttt{Cofix $f$}~'' can be used to do that. Similarly to the tactic \texttt{fix} indicated in Section \ref{WellFoundedRecursion}, this tactic introduces an extra hypothesis $f$ into the context, whose type is the same as the current goal. Note that the applications of $f$ in the proof \textsl{must be guarded}. In order to prevent us from doing unguarded calls, we can define a tactic that always apply a constructor before using $f$ \refmancite{Chapter \ref{WritingTactics}} : \begin{alltt} Ltac infiniteproof f := cofix f; constructor; [clear f| simpl; try (apply f; clear f)]. \end{alltt} In the example above, this tactic produces a much simpler proof that the former one: \begin{alltt} Theorem map_iterate' : {\prodsym} ((A:Set)f:A{\arrow}A)(x:A), EqSt (iterate f (f x)) (map f (iterate f x)). Proof. infiniteproof map_iterate'. reflexivity. Qed. \end{alltt} \begin{exercise} Define a co-inductive type $Nat$ containing non-standard natural numbers --this is, verifying $$\exists m \in \mbox{\texttt{Nat}}, \forall\, n \in \mbox{\texttt{Nat}}, n Infinite l A : Set a : A l : LList A H0 : ~ Finite (LCons a l) ============================ Infinite l \end{alltt} At this point, one must not apply \citecoq{H}! . It would be possible to solve the current goal by an inversion of ``~\citecoq{Finite (LCons a l)}~'', but, since the guard condition would be violated, the user would get an error message after typing \citecoq{Qed}. In order to satisfy the guard condition, we apply the constructor of \citecoq{Infinite}, \emph{then} apply \citecoq{H}. \begin{alltt} constructor. apply H. red; intro H1;case H0. constructor. trivial. Qed. \end{alltt} The reader is invited to replay this proof and understand each of its steps. \bibliographystyle{abbrv} \bibliography{manbiblio,morebib} \end{document}